**Physics Problem on Circular Motion: Roller Coaster in a Vertical Dip**---**Problem 7:**A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius \( r \). A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If \( r = 21.5 \) m, how fast is the roller coaster traveling at the bottom of the dip?\[ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \text{m/s} \]--- **Detailed Explanation:**To solve this problem, we need to use principles related to circular motion and forces:1. **Centripetal Force Requirement:** At the bottom of the dip, the net force acting on the passenger provides the centripetal force required to keep her moving in a circular path. The centripetal force (\( F_c \)) is given by: \[ F_c = \frac{mv^2}{r} \] where \( m \) is the mass of the passenger, \( v \) is the speed of the roller coaster at the bottom of the dip, and \( r \) is the radius of the vertical circle.2. **Perceived Weight and Real Weight Relationship:** The passenger feels a force exerted by the seat that is five times her real weight. This force includes the real weight plus the centripetal force. So: \[ 5mg = mg + F_c \]3. **Substituting \( F_c \):** \[ 5mg = mg + \frac{mv^2}{r} \]4. **Solving for \( v \):** - Rearrange the equation to isolate \( \frac{mv^2}{r} \): \[ 5mg - mg = \frac{mv^2}{r} \] \[ 4mg = \frac{mv^2}{r} \] - Cancel \( m \) from both sides (assuming \( m \neq 0 \)): \[ 4g = \frac{v^2}{r} \] - Multiply both sides by \( r \):

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the at of the car pushing upward on her with a force equal to five times her as she goes through the dip. If r = 21.5 how fast is the roller coaster traveling at the bottom of the dip? m/s sf60 f60 st O ssfo

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the at of the car pushing upward on her with a force equal to five times her as she goes through the dip. If r = 21.5 how fast is the roller coaster traveling at the bottom of the dip? m/s sf60 f60 st O ssfo

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

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**Physics Problem on Circular Motion: Roller Coaster in a Vertical Dip**

---

**Problem 7:**

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius \( r \). A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If \( r = 21.5 \) m, how fast is the roller coaster traveling at the bottom of the dip?

\[ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \text{m/s} \]

---

**Detailed Explanation:**

To solve this problem, we need to use principles related to circular motion and forces:

1. **Centripetal Force Requirement:**

At the bottom of the dip, the net force acting on the passenger provides the centripetal force required to keep her moving in a circular path. The centripetal force (\( F_c \)) is given by:

\[ F_c = \frac{mv^2}{r} \]

where \( m \) is the mass of the passenger, \( v \) is the speed of the roller coaster at the bottom of the dip, and \( r \) is the radius of the vertical circle.

2. **Perceived Weight and Real Weight Relationship:**

The passenger feels a force exerted by the seat that is five times her real weight. This force includes the real weight plus the centripetal force. So:

\[ 5mg = mg + F_c \]

3. **Substituting \( F_c \):**

\[ 5mg = mg + \frac{mv^2}{r} \]

4. **Solving for \( v \):**

- Rearrange the equation to isolate \( \frac{mv^2}{r} \):

\[ 5mg - mg = \frac{mv^2}{r} \]
\[ 4mg = \frac{mv^2}{r} \]

- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ 4g = \frac{v^2}{r} \]

- Multiply both sides by \( r \):

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